EE 131A
Problem Set 6
Probability
Wednesday, November 10, 2010
Instructor: Lara Dolecek
Due: Wednesday, November 17, 2010
Reading: Chapter 5 of
Probability, Statistics, and Random Processes
by A. LeonGarcia
100 points total
1.
Coin tossing and two RVs.
Problem 5.2, page 288 of ALG
Solution
: (a) The underlying space
S
=
{
HHHH, HHHT, HHTH, HHTT, HTHH,
HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
}
.
Let
W
and
Z
be the number of heads obtained by Carlos and Michael respectively when
each ﬂip a fair coin twice, then the range of the pair (
W,Z
) is given by:
S
WZ
=
{
(0
,
0)
,
(0
,
1)
,
(0
,
2)
,
(1
,
0)
,
(1
,
1)
,
(1
,
2)
,
(2
,
0)
,
(2
,
1)
,
(2
,
2)
}
The mapping between (
w,z
) and (
x,y
) can be done as
x
=
w

z
and
y
=
w
+
z
,
therefore the range of the pair (
X,Y
) is given by:
S
XY
=
{
(0
,
0)
,
(

1
,
1)
,
(

2
,
2)
,
(1
,
1)
,
(0
,
2)
,
(

1
,
3)
,
(2
,
2)
,
(1
,
3)
,
(0
,
4)
}
(b) The probabilities for all values of (
X,Y
) are given by:
P
[
X
= 0
,Y
= 0] =
P
[
W
= 0
,Z
= 0] =
P
[
W
= 0]
P
[
Z
= 0] = 1
/
16
P
[
X
=

1
,Y
= 1] =
P
[
W
= 0
,Z
= 1] =
P
[
W
= 0]
P
[
Z
= 1] = 1
/
8
P
[
X
=

2
,Y
= 2] =
P
[
W
= 0
,Z
= 2] =
P
[
W
= 0]
P
[
Z
= 2] = 1
/
16
P
[
X
= 1
,Y
= 1] =
P
[
W
= 1
,Z
= 0] =
P
[
W
= 1]
P
[
Z
= 0] = 1
/
8
P
[
X
= 0
,Y
= 2] =
P
[
W
= 1
,Z
= 1] =
P
[
W
= 1]
P
[
Z
= 1] = 1
/
4
P
[
X
=

1
,Y
= 3] =
P
[
W
= 1
,Z
= 2] =
P
[
W
= 1]
P
[
Z
= 2] = 1
/
8
P
[
X
= 2
,Y
= 2] =
P
[
W
= 2
,Z
= 0] =
P
[
W
= 2]
P
[
Z
= 0] = 1
/
16
P
[
X
= 1
,Y
= 3] =
P
[
W
= 2
,Z
= 1] =
P
[
W
= 2]
P
[
Z
= 1] = 1
/
8
P
[
X
= 0
,Y
= 4] =
P
[
W
= 2
,Z
= 2] =
P
[
W
= 2]
P
[
Z
= 2] = 1
/
16
(c) Based on the above probabilities, we have:
P
[
X
+
Y
= 1] = 0
P
[
X
+
Y
= 2] =
P
[
X
= 1
,Y
= 1] +
P
[
X
= 0
,Y
= 2] +
P
[
X
=

1
,Y
= 3] = 1
/
2
2.
Coin tossing and two RVs and pdfs.
Problem 5.10, page 289 of ALG